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3 years ago

Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:

Physics

1 answer:

Deffense [45]3 years ago

6 0

Answer:

a) $T=0.01s$

b) $T=0.001s$

c) $T=0.00001s$

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

$T=\frac{1}{f}$

Therefore

For

$T=100 Hz$

$T=\frac{1}{100}$

$T=0.01s$

For

$F=1kHz$

$T=\frac{1}{1000}$

$T=0.001s$

For

$F=100kHz$

$T=\frac{1}{100*100}$

$T=0.00001s$

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How much energy is needed to heat and melt 3.0 kg of copper initially at 83°C?

Ne4ueva [31]

As we know that in order to melt the copper we need to take the temperature of copper to its melting point

So here heat required to raise the temperature of copper is given as

$Q = ms\Delta T$

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now we have

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now in order to melt the copper we know the heat required is

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here we know that

L = 205 kJ/kg

now from above formula

$Q = 3(205) kJ$

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now total heat required will be

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3 years ago

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satela [25.4K]

Answer:

Nuclear medicine or PET scanning (Positron Emission Tomography)

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2 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

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3 years ago

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quester [9]

Answer:

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Then acceleration a = F/m

a = 42/0.30

a = 140 m/s

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